The Hook: From Refrigerators to Anesthesia
Ever wondered what makes your refrigerator cool? The answer lies in halogenated compounds! Refrigerants like R-22 (CHClF₂) are alkyl halides. Even in medicine, halothane (CF₃-CHClBr) was widely used as an anesthetic. The tear gas used for riot control? That’s chloroacetophenone - another halogen compound!
Here’s the JEE question: Why are alkyl halides so reactive compared to their parent alkanes? And why does iodoethane react faster than fluoroethane in nucleophilic substitution?
The Core Concept
What are Alkyl Halides?
Alkyl halides (haloalkanes) are compounds where one or more hydrogen atoms of an alkane are replaced by halogen atoms (F, Cl, Br, I).
General formula:
$$\boxed{\text{R-X}}$$where R = alkyl group, X = F, Cl, Br, or I
Why Alkyl Halides are Reactive
The key to understanding alkyl halide chemistry is the C-X bond polarity:
$$\boxed{\delta^+ \text{C} - \delta^- \text{X}}$$Electronegativity values:
- Carbon: 2.5
- Fluorine: 4.0, Chlorine: 3.0, Bromine: 2.8, Iodine: 2.5
This creates a polar covalent bond making carbon electrophilic (electron-deficient) and susceptible to nucleophilic attack!
Classification of Alkyl Halides
1. Based on Number of Halogen Atoms
Monohalides: One halogen atom
- Example: CH₃Br (methyl bromide)
Dihalides: Two halogen atoms
- Geminal (gem): Both halogens on same carbon
- Example: CH₃-CHCl₂ (1,1-dichloroethane)
- Vicinal (vic): Halogens on adjacent carbons
- Example: CH₂Cl-CH₂Cl (1,2-dichloroethane)
Trihalides, Tetrahalides: Three or four halogens
- Example: CHCl₃ (chloroform), CCl₄ (carbon tetrachloride)
2. Based on Hybridization of Carbon Bearing Halogen
sp³ Hybridized: Alkyl halides (saturated)
- Example: CH₃CH₂Cl
sp² Hybridized: Vinyl and aryl halides (unsaturated)
- Vinyl: CH₂=CHCl
- Aryl: C₆H₅Cl (chlorobenzene)
sp Hybridized: Very rare
- Example: HC≡C-Cl
3. Based on Carbon Position (Most Important for JEE)
Primary (1°) Alkyl Halide:
$$\boxed{\text{R-CH}_2\text{-X}}$$Halogen attached to primary carbon (carbon bonded to one other carbon)
Example: CH₃CH₂Br (ethyl bromide)
Secondary (2°) Alkyl Halide:
$$\boxed{\text{R}_2\text{CH-X}}$$Halogen attached to secondary carbon (carbon bonded to two other carbons)
Example: (CH₃)₂CHBr (isopropyl bromide)
Tertiary (3°) Alkyl Halide:
$$\boxed{\text{R}_3\text{C-X}}$$Halogen attached to tertiary carbon (carbon bonded to three other carbons)
Example: (CH₃)₃CBr (tert-butyl bromide)
“Count the neighbors of the carbon with X”
- 1 neighbor = Primary (1°)
- 2 neighbors = Secondary (2°)
- 3 neighbors = Tertiary (3°)
JEE Strategy: This classification determines reaction mechanism (SN1 vs SN2)!
Interactive Demo: Visualize Alkyl Halide Classifications
Explore primary, secondary, and tertiary alkyl halide 3D structures.
IUPAC Nomenclature
Rules for Naming Alkyl Halides
1. Select longest carbon chain containing the C-X bond
2. Number the chain from the end giving the halogen the lowest number
3. Name halogen as substituent using prefixes:
- F → fluoro
- Cl → chloro
- Br → bromo
- I → iodo
4. Arrange multiple substituents alphabetically (not by number)
5. For multiple identical halogens use di-, tri-, tetra- prefixes
Examples with Solutions
Structure: CH₃-CH₂-CH₂-Br
Step-by-step:
- Longest chain: 3 carbons = propane
- Number from end nearest Br: Br is at C-1
- Name: 1-bromopropane
Common name: n-propyl bromide
Structure:
CH₃
|
CH₃—CH—CH—CH₃
|
Br
Step-by-step:
- Longest chain: 4 carbons = butane
- Substituents: Br at C-2, CH₃ at C-3
- Name: 3-methyl-2-bromobutane
Note: NOT 2-bromo-3-methylbutane (alphabetical order: b before m)
Structure: CH₂Cl-CHCl-CH₂Cl
Solution:
- Three carbons = propane
- Three Cl atoms at C-1, C-2, C-3
- Name: 1,2,3-trichloropropane
Not: trichloropropane (must specify positions)
Structure:
F
|
CH₃—C—CH₃
|
Cl
Solution:
- Three carbons = propane
- C-2 has both F and Cl
- Alphabetical: chloro before fluoro
- Name: 2-chloro-2-fluoropropane
Common names are still used in JEE questions!
| IUPAC Name | Common Name |
|---|---|
| Chloromethane | Methyl chloride |
| 2-chloropropane | Isopropyl chloride |
| 2-chloro-2-methylpropane | tert-butyl chloride |
| Chloroethene | Vinyl chloride |
| Chlorobenzene | Phenyl chloride |
JEE Tip: Know both nomenclatures - questions use them interchangeably!
Preparation of Alkyl Halides
Method 1: From Alcohols (Most Important for JEE)
General Reaction:
$$\boxed{\text{R-OH} + \text{HX} \rightarrow \text{R-X} + \text{H}_2\text{O}}$$Using Hydrogen Halides (HX)
Reactivity order of alcohols:
$$\boxed{3° > 2° > 1°}$$Reactivity order of hydrogen halides:
$$\boxed{\text{HI} > \text{HBr} > \text{HCl}}$$Why?
- 3° alcohols form more stable carbocations
- HI is strongest acid, best leaving group
Example:
$$(CH_3)_3C-OH + HCl \xrightarrow{\text{conc. HCl}} (CH_3)_3C-Cl + H_2O$$“Tertiary alcohols are Terrific, react Terrifically fast!”
For alcohols: 3° > 2° > 1° (follows carbocation stability)
For HX: “I am Brave but Clumsy with Fluorine”
- HI > HBr > HCl (HF doesn’t work well)
Using Phosphorus Halides
With PCl₃ (for R-Cl):
$$\boxed{3\text{R-OH} + \text{PCl}_3 \rightarrow 3\text{R-Cl} + \text{H}_3\text{PO}_3}$$With PCl₅ (for R-Cl):
$$\boxed{\text{R-OH} + \text{PCl}_5 \rightarrow \text{R-Cl} + \text{POCl}_3 + \text{HCl}}$$With PBr₃ (for R-Br):
$$\boxed{3\text{R-OH} + \text{PBr}_3 \rightarrow 3\text{R-Br} + \text{H}_3\text{PO}_3}$$With PI₃ (for R-I):
$$\boxed{3\text{R-OH} + \text{PI}_3 \rightarrow 3\text{R-I} + \text{H}_3\text{PO}_3}$$Advantage: Works well with all types of alcohols (1°, 2°, 3°)
Using Thionyl Chloride (SOCl₂)
$$\boxed{\text{R-OH} + \text{SOCl}_2 \rightarrow \text{R-Cl} + \text{SO}_2 + \text{HCl}}$$Advantage: Gaseous byproducts (SO₂ and HCl) escape, leaving pure product!
Example:
$$\text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow$$Q: Which reagent gives the purest alkyl chloride from an alcohol?
Options: (A) Conc. HCl (B) PCl₃ (C) PCl₅ (D) SOCl₂
Answer: (D) SOCl₂
Reason: Gaseous byproducts (SO₂ and HCl) escape automatically, no separation needed!
Method 2: Halogenation of Alkanes
Free radical substitution:
$$\boxed{\text{R-H} + \text{X}_2 \xrightarrow{h\nu \text{ or heat}} \text{R-X} + \text{HX}}$$Mechanism: See Alkanes for detailed free radical mechanism
Disadvantage: Mixture of products (poly-substituted)
Example:
$$\text{CH}_4 + \text{Cl}_2 \xrightarrow{h\nu} \text{CH}_3\text{Cl} + \text{CH}_2\text{Cl}_2 + \text{CHCl}_3 + \text{CCl}_4$$Wrong thinking: “Halogenation gives pure monohalide”
Reality: Free radical halogenation gives mixture of products
JEE Tip: This method is NOT preferred for preparation. It’s important for understanding mechanism, not synthesis!
Method 3: Addition of HX to Alkenes
Markovnikov’s Addition:
$$\boxed{\text{R-CH=CH}_2 + \text{HX} \rightarrow \text{R-CHX-CH}_3}$$Markovnikov’s Rule: Hydrogen adds to carbon with more hydrogens
Example:
$$\text{CH}_3\text{-CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{-CHBr-CH}_3$$Anti-Markovnikov Addition (Peroxide Effect):
$$\boxed{\text{R-CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{R-CH}_2\text{-CH}_2\text{Br}}$$Note: Peroxide effect works ONLY with HBr, not HCl or HI!
“HBr is Brave with peroxide, HCl is Coward, HI is Indifferent”
Peroxide effect: Only HBr undergoes anti-Markovnikov addition
Why?
- HCl: C-Cl bond too strong to break
- HI: I• radical too unstable
- HBr: Just right! (Goldilocks principle)
Related: Alkenes
Method 4: Halogen Exchange (Finkelstein Reaction)
For making R-I from R-Cl or R-Br:
$$\boxed{\text{R-X} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaX}}$$Driving force: NaCl and NaBr are insoluble in acetone, precipitate out
Example:
$$\text{CH}_3\text{Br} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{I} + \text{NaBr} \downarrow$$Method 5: Swarts Reaction
For making alkyl fluorides:
$$\boxed{\text{R-X} + \text{AgF (or Hg}_2\text{F}_2\text{)} \rightarrow \text{R-F} + \text{AgX}}$$Example:
$$\text{CH}_3\text{Br} + \text{AgF} \rightarrow \text{CH}_3\text{F} + \text{AgBr}$$Method 6: Hunsdiecker Reaction
From carboxylic acid salts:
$$\boxed{\text{R-COOAg} + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{R-Br} + \text{CO}_2 + \text{AgBr}}$$Note: Product has ONE carbon less than starting material!
Example:
$$\text{CH}_3\text{CH}_2\text{COOAg} + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{CO}_2 + \text{AgBr}$$Q: How will you prepare 1-bromopropane from butan-1-ol?
Solution:
Step 1: Convert butan-1-ol to butanoic acid
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$$Step 2: Convert to silver salt
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{AgNO}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOAg}$$Step 3: Hunsdiecker reaction
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOAg} + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{CO}_2$$Answer: 1-bromopropane (one carbon less than starting alcohol!)
Physical Properties
1. Nature of C-X Bond
Bond characteristics:
- Polar covalent (C⁺-X⁻)
- Dipole moment decreases: CH₃F > CH₃Cl > CH₃Br > CH₃I
Why?
- Electronegativity decreases down the group
- But bond length increases, offsetting polarity
2. Boiling Points
Trend in same alkyl group:
$$\boxed{\text{R-I} > \text{R-Br} > \text{R-Cl} > \text{R-F}}$$Why?
- Molecular mass increases: I > Br > Cl > F
- van der Waals forces increase with mass
- Higher intermolecular forces → higher boiling point
Trend with carbon chain:
$$\boxed{\text{Longer chain} \rightarrow \text{Higher boiling point}}$$Trend with branching:
$$\boxed{\text{More branching} \rightarrow \text{Lower boiling point}}$$Example:
- 1-bromobutane: BP = 101°C (straight chain)
- 2-bromo-2-methylpropane: BP = 73°C (branched)
3. Density
All alkyl halides are denser than water (except some fluorides)
Trend:
$$\boxed{\text{R-I} > \text{R-Br} > \text{R-Cl} > \text{R-F}}$$Example: CHCl₃ (chloroform) has density = 1.48 g/mL (water = 1.0 g/mL)
4. Solubility
In water: Slightly soluble or insoluble
- C-X bond is polar, but not enough to break H-bonds in water
- Larger alkyl groups → less soluble
In organic solvents: Highly soluble
- “Like dissolves like” principle
- Non-polar alkyl group compatible with organic solvents
Reactivity of C-X Bond
Bond Strength
Bond dissociation energy (kJ/mol):
| Bond | Energy |
|---|---|
| C-F | 485 |
| C-Cl | 339 |
| C-Br | 285 |
| C-I | 213 |
Trend:
$$\boxed{\text{C-F} > \text{C-Cl} > \text{C-Br} > \text{C-I}}$$Reactivity in Nucleophilic Substitution
Reactivity order:
$$\boxed{\text{R-I} > \text{R-Br} > \text{R-Cl} > \text{R-F}}$$Why the paradox?
- Though C-I bond is weakest, I⁻ is best leaving group!
- C-F bond is strongest, F⁻ is poor leaving group
Leaving group ability depends on:
Stability of X⁻ after leaving
- I⁻ is most stable (large, charge distributed)
- F⁻ is least stable (small, concentrated charge)
Bond strength (secondary factor)
- Weaker bond → easier to break
Best leaving groups (most stable anions):
$$\boxed{\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-}$$This determines reactivity in SN1 and SN2 reactions!
Related: SN1 and SN2 Reactions
Common Mistakes to Avoid
Wrong: “C-F bond is strongest, so R-F is most reactive”
Correct: C-F bond is strongest, so R-F is LEAST reactive!
- Stronger bond → harder to break
- F⁻ is poor leaving group
JEE Tip: Reactivity = how easily bond breaks AND how stable the leaving group is
Wrong: “2-methyl-3-chlorobutane” (giving methyl lower number)
Correct: “3-chloro-2-methylbutane” (alphabetical: c before m)
JEE Rule: Alphabetical order for different substituents, NOT numerical order!
Wrong: Using free radical halogenation to prepare specific alkyl halide
Correct: Free radical halogenation gives mixture; use alcohol + HX for specific product
JEE Strategy:
- From alcohol → use HX, PCl₅, or SOCl₂
- From alkene → use HX (Markovnikov)
- For specific position → choose starting material carefully
Practice Problems
Level 1: Foundation (NCERT)
Q: Name the following compounds using IUPAC nomenclature:
(a) CH₃-CH₂-CH(Cl)-CH₃ (b) (CH₃)₂CH-CH₂Br (c) CH₂Cl-CH₂Cl
Solutions:
(a) 2-chlorobutane
- 4 carbons = butane
- Cl at position 2
(b) 1-bromo-2-methylpropane
- Longest chain: 3 carbons = propane
- Br at C-1, CH₃ at C-2
(c) 1,2-dichloroethane
- 2 carbons = ethane
- Two Cl atoms at C-1 and C-2
Q: Classify as 1°, 2°, or 3° alkyl halide:
(a) (CH₃)₂CH-Br (b) CH₃CH₂CH₂Cl (c) (CH₃)₃C-I
Solutions:
(a) Secondary (2°) - Br attached to carbon with 2 alkyl groups
(b) Primary (1°) - Cl attached to carbon with 1 alkyl group
(c) Tertiary (3°) - I attached to carbon with 3 alkyl groups
Level 2: JEE Main
Q: Arrange in increasing order of boiling point: CH₃CH₂Br, CH₃Br, CH₃CH₂CH₂Br
Solution:
Factors: All are bromides, so compare chain length
- More carbons → higher molecular mass → higher BP
Order: CH₃Br < CH₃CH₂Br < CH₃CH₂CH₂Br
Reasoning:
- CH₃Br: 1 carbon, lowest mass
- CH₃CH₂Br: 2 carbons, medium mass
- CH₃CH₂CH₂Br: 3 carbons, highest mass
Q: How will you prepare 2-bromopropane from propan-2-ol?
Solution:
Method 1: Using HBr
$$\text{CH}_3\text{-CHOH-CH}_3 + \text{HBr} \rightarrow \text{CH}_3\text{-CHBr-CH}_3 + \text{H}_2\text{O}$$Method 2: Using PBr₃
$$3\text{CH}_3\text{-CHOH-CH}_3 + \text{PBr}_3 \rightarrow 3\text{CH}_3\text{-CHBr-CH}_3 + \text{H}_3\text{PO}_3$$Best method: HBr (since it’s 2° alcohol, reacts well with HBr)
Level 3: JEE Advanced
Q: How will you convert ethanol to 1-iodoethane? Give all reagents.
Solution:
Method 1: Direct (using HI)
$$\text{CH}_3\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O}$$Method 2: Via Finkelstein
Step 1: Make 1-bromoethane
$$\text{CH}_3\text{CH}_2\text{OH} + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$$Step 2: Halogen exchange
$$\text{CH}_3\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{CH}_2\text{I} + \text{NaBr}$$Method 3: Using PI₃
$$3\text{CH}_3\text{CH}_2\text{OH} + \text{PI}_3 \rightarrow 3\text{CH}_3\text{CH}_2\text{I} + \text{H}_3\text{PO}_3$$Best answer: Method 1 or 3 (direct methods)
Q: Arrange in order of reactivity toward nucleophilic substitution: 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane
Solution:
Structure identification:
- 1-bromobutane: CH₃CH₂CH₂CH₂Br (1°)
- 2-bromobutane: CH₃CHBrCH₂CH₃ (2°)
- 2-bromo-2-methylpropane: (CH₃)₃CBr (3°)
For SN2 reaction:
$$\boxed{1° > 2° > 3°}$$Order: 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane
For SN1 reaction:
$$\boxed{3° > 2° > 1°}$$Order: 2-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane
JEE Tip: Question should specify SN1 or SN2! If not specified, assume SN2.
Related: SN1 and SN2 Mechanisms
Quick Revision Box
| Topic | Key Points | JEE Formula/Rule |
|---|---|---|
| Classification | 1°, 2°, 3° based on carbon | Count carbons attached to C-X |
| From Alcohols | R-OH → R-X | HX, PCl₅, PBr₃, SOCl₂ |
| Best for R-Cl | Purest product | SOCl₂ (gaseous byproducts) |
| From Alkenes | Addition of HX | Markovnikov (or anti with peroxide) |
| Finkelstein | R-Cl → R-I | NaI in acetone |
| Swarts | R-Cl → R-F | AgF or Hg₂F₂ |
| Boiling Point | R-I > R-Br > R-Cl > R-F | Due to molecular mass |
| Reactivity | R-I > R-Br > R-Cl > R-F | Leaving group ability |
| Nomenclature | Alphabetical order | chloro, bromo, fluoro, iodo |
Connection to Other Topics
Prerequisites:
- Alkanes - Parent hydrocarbons, halogenation mechanism
- Alcohols - Conversion to alkyl halides
- Organic Principles - IUPAC nomenclature
Next Topics:
- SN1 and SN2 Reactions - Nucleophilic substitution mechanisms
- Elimination Reactions - E1 and E2 mechanisms
- Haloarenes - Aromatic halides
Applications:
- Grignard Reagents - R-X + Mg → RMgX
- Nucleophilic Substitution - Making amines from R-X
Teacher’s Summary
1. Alkyl Halides = R-X (polar C-X bond makes carbon electrophilic)
2. Classification by carbon type determines reactivity:
- 1° → SN2 preferred
- 3° → SN1 preferred
3. Preparation Methods (High-yield for JEE):
- From alcohols: HX, PCl₅, PBr₃, SOCl₂ (best for R-Cl)
- From alkenes: HX addition (Markovnikov or anti-Markovnikov)
- Finkelstein: R-X → R-I (using NaI/acetone)
4. Physical Properties:
- Boiling point: R-I > R-Br > R-Cl > R-F (molecular mass)
- Denser than water
- Slightly soluble in water, very soluble in organic solvents
5. Reactivity Order:
- R-I > R-Br > R-Cl > R-F (leaving group ability)
- Opposite to bond strength!
6. JEE Strategy:
- Know BOTH IUPAC and common names
- Choose preparation based on starting material
- Understand 1°/2°/3° classification for mechanism questions
“The C-X bond polarity is the key - it makes carbon electron-deficient and reactive toward nucleophiles!”
Next, master SN1 and SN2 mechanisms to understand how alkyl halides undergo substitution reactions!